The power-ful problem

P=ma.v

Given x=kt^p

So, a=ckt^(p-2) c=p(p-1) v=pkt^(p-1)

 ma.v=p=constant (power is contant)

Hence       t^(p-1+p-2) = 1

2p-3=0 , p= 1.5

From the definition of Power in Mechanics, we can write the relation that:- $P = Fv$ $P = mav$ Using the relation that $a = v\dfrac{dv}{dx}$ . $P = mv\dfrac{dv}{dx}v$ $P = mv^{2}\dfrac{dv}{dx}$ Integrating the above expression we get:- $\int_{0}^{x} Pdx = \int_{0}^{v} mv^{2}dv$ $Px = \dfrac{mv^{3}}{3}$ $\dfrac{3Px}{m} = v^{3}$ $\dfrac{dx}{dt} = \Big( \dfrac{3Px}{m}\Big) ^{\frac{1}{3}}$ $\dfrac{dx}{x^{\frac{1}{3}}} = \Big( \dfrac{3P}{m} \Big)^{\frac{1}{3}} dt$ Again integrating both sides we get:- $\int_{0}^{x}\dfrac{dx}{x^{\frac{1}{3}}} = \int_{0}^{t} \Big( \dfrac{3P}{m}\Big)^{\frac{1}{3}} dt$ $\dfrac{3x^{\frac{2}{3}}}{2} = \Big( \dfrac{3P}{m}\Big)^{\frac{1}{3}} t$ Now cubing both sides:- $\dfrac{27x^{2}}{8} = \dfrac{3Pmt^{3}}{m}$ $x^{2} = \dfrac{8Pt^{3}}{9m}$ $x = \dfrac{2\sqrt{2}}{3} \sqrt{\dfrac{P}{m}}t\sqrt{t}$ Hence we can write that:- $x = kt^{1.5}$